Divisibility and Modular Arithmetic Section 4 1 Division

Divisibility and Modular Arithmetic Section 4. 1

Division Definition: If a and b are integers with a ≠ 0, then b if there exists an integer c such that b = ac. a divides • When a divides b we say that a is a factor or divisor of b and that b is a multiple of a. • The notation a | b denotes that a divides b. • If a | b, then b/a is an integer. • If a does not divide b, we write a ∤ b. Example: Determine whether 3 | 7 and whether 3 | 12.

Division “Algorithm” • When an integer is divided by a positive integer, there is a quotient and a remainder. This is traditionally called the “Division Algorithm, ” but is really a theorem. Division Algorithm: If a is an integer and d a positive integer, then there are unique integers q and r, with 0 ≤ r < d, such that a = dq + r (proved in Section 5. 2). • • d is called the divisor. a is called the dividend. q is called the quotient. r is called the remainder. Definitions of Functions div and mod q = a div d r = a mod d

Division “Algorithm” • Examples: • What are the quotient and remainder when 101 is divided by 11? Solution: The quotient when 101 is divided by 11 is 9 = 101 div 11, and the remainder is 2 = 101 mod 11. • What are the quotient and remainder when − 11 is divided by 3? Solution: The quotient when − 11 is divided by 3 is − 4 = − 11 div 3, and the remainder is 1 = − 11 mod 3.

Congruence Relation Definition: If a and b are integers and m is a positive integer, then a is congruent to b modulo m if m divides a – b. • The notation a ≡ b (mod m) says that a is congruent to b modulo m. • We say that a ≡ b (mod m) is a congruence and that m is its modulus. • Two integers are congruent mod m if and only if they have the same remainder when divided by m. • If a is not congruent to b modulo m, we write a ≢ b (mod m)

Congruence Relation Example: Determine whether 17 is congruent to 5 modulo 6 and whether 24 and 14 are congruent modulo 6. Solution: • 17 ≡ 5 (mod 6) because 6 divides 17 − 5 = 12. • 24 ≢ 14 (mod 6) since 6 divides 24 − 14 = 10 is not divisible by 6.

More on Congruences Theorem 4: Let m be a positive integer. The integers a and b are congruent modulo m if and only if there is an integer k such that a = b + km. Proof: • If a ≡ b (mod m), then (by the definition of congruence) m | a – b. Hence, there is an integer k such that a – b = km and equivalently a = b + km. • Conversely, if there is an integer k such that a = b + km, then km = a – b. Hence, m | a – b and a ≡ b (mod m).

The Relationship between (mod m) and mod m Notations • The use of “mod” in a ≡ b (mod m) and a mod m = b are different. • a ≡ b (mod m) is a relation on the set of integers. • In a mod m = b, the notation mod denotes a function. • The relationship between these notations is made clear in this theorem. • Theorem 3: Let a and b be integers, and let m be a positive integer. Then a ≡ b (mod m) if and only if a mod m = b mod m. (Proof in the exercises)

Congruences of Sums and Products Theorem 5: Let m be a positive integer. If a ≡ b (mod m) and c ≡ d (mod m), then a + c ≡ b + d (mod m) and ac ≡ bd (mod m) Proof: • Because a ≡ b (mod m) and c ≡ d (mod m), by Theorem 4 there are integers s and t with b = a + sm and d = c + tm. • Therefore, • b + d = (a + sm) + (c + tm) = (a + c) + m(s + t) and • b d = (a + sm) (c + tm) = ac + m(at + cs + stm). • Hence, a + c ≡ b + d (mod m) and ac ≡ bd (mod m). Example: Because 7 ≡ 2 (mod 5) and 11 ≡ 1 (mod 5) , it follows from Theorem 5 that 18 = 7 + 11 ≡ 2 + 1 = 3 (mod 5) 77 = 7 11 ≡ 2 + 1 = 3 (mod 5)

Algebraic Manipulation of Congruences • Multiplying both sides of a valid congruence by an integer preserves validity. If a ≡ b (mod m) holds then c∙a ≡ c∙b (mod m), where c is any integer, holds by Theorem 5 with d = c. • Adding an integer to both sides of a valid congruence preserves validity. If a ≡ b (mod m) holds then c + a ≡ c + b (mod m), where c is any integer, holds by Theorem 5 with d = c. • Dividing a congruence by an integer does not always produce a valid congruence. Example: The congruence 14≡ 8 (mod 6) holds. But dividing both sides by 2 does not produce a valid congruence since 14/2 = 7 and 8/2 = 4, but 7≢ 4 (mod 6).

Computing the mod m Function of Products and Sums • We use the following corollary to Theorem 5 to compute the remainder of the product or sum of two integers when divided by m from the remainders when each is divided by m. Corollary: Let m be a positive integer and let a and b be integers. Then (a + b) (mod m) = ((a mod m) + (b mod m)) mod m and ab mod m = ((a mod m) (b mod m)) mod m. (proof in text)

Arithmetic Modulo m Definitions: Let Zm be the set of nonnegative integers less than m: {0, 1, …. , m− 1} • The operation +m is defined as a +m b = (a + b) mod m. This is addition modulo m. • The operation ∙m is defined as a ∙m b = (ab) mod m. This is multiplication modulo m. • Using these operations is said to be doing arithmetic modulo m. Example: Find 7 +11 9 and 7 ∙ 11 9. Solution: Using the definitions above: • 7 +11 9 = (7 + 9) mod 11 = 16 mod 11 = 5 • 7 ∙ 11 9 = (7 ∙ 9) mod 11 = 63 mod 11 = 8

Arithmetic Modulo m • The operations +m and ∙m satisfy many of the same properties as ordinary addition and multiplication. • Closure: If a and b belong to Zm , then a +m b and a ∙m b belong to Zm. • Associativity: If a, b, and c belong to Zm , then (a +m b) +m c = a +m (b +m c) and (a ∙m b) ∙m c = a ∙m (b ∙m c). • Commutativity: If a and b belong to Zm , then a +m b = b +m a and a ∙m b = b ∙m a. • Identity elements: The elements 0 and 1 are identity elements for addition and multiplication modulo m, respectively. • If a belongs to Zm , then a +m 0 = a and a ∙m 1 = a. continued →

Arithmetic Modulo m • Additive inverses: If a≠ 0 belongs to Zm , then m− a is the additive inverse of a modulo m and 0 is its own additive inverse. • a +m (m− a ) = 0 and 0 +m 0 = 0 • Distributivity: If a, b, and c belong to Zm , then • a ∙m (b +m c) = (a ∙m b) +m (a ∙m c) and (a +m b) ∙m c = (a ∙m c) +m (b ∙m c). • Exercises 42 -44 ask for proofs of these properties. • Multiplicative inverses have not been included since they do not always exist. For example, there is no multiplicative inverse of 2 modulo 6.

Modular Exponentiation • Efficient algorithms use the binary expansion of the expansion • Any number can be represented in binary form • Therefore, any power, xn, can be represented as products of x 1 , x 2 , x 4 , x 8 , … • Example: x 13 = x 8+4+1 = x 8 x 4 x • For modular exponentiation: • Example: x 13 (mod b) = [x 8(mod b)⋅x 4(mod b)⋅x(mod b)] (mod b)

Exponentiation Algorithm // Inputs: “base” and “power” int prod = 1; // running product int val = base; int n = power; while (n > 0) { if (n & 1) prod = prod*val; n >>= 1 // Same as n *= 2 val = val*val; } // val = base ^ power

Integer Representations and Algorithms Section 4. 2

Representations of Integers • In the modern world, we use decimal, or base 10, notation to represent integers. For example when we write 965, we mean 9∙ 102 + 6∙ 101 + 5∙ 100. • We can represent numbers using any base b, where b is a positive integer greater than 1. • The bases b = 2 (binary), b = 8 (octal) , and b= 16 (hexadecimal) are important for computing and communications • The ancient Mayans used base 20 and the ancient Babylonians used base 60.

Base b Representations • We can use positive integer b greater than 1 as a base, because of this theorem: Theorem 1: Let b be a positive integer greater than 1. Then if n is a positive integer, it can be expressed uniquely in the form: n = akbk + ak-1 bk-1 + …. + a 1 b + a 0 where k is a nonnegative integer, a 0, a 1, …. ak are nonnegative integers less than b, and ak≠ 0. The aj, j = 0, …, k are called the base-b digits of the representation. (We will prove this using mathematical induction in Section 5. 1. ) • The representation of n given in Theorem 1 is called the base b expansion of n and is denoted by (akak-1…. a 1 a 0)b. • We usually omit the subscript 10 for base 10 expansions.

Binary Expansions Most computers represent integers and do arithmetic with binary (base 2) expansions of integers. In these expansions, the only digits used are 0 and 1. Example: What is the decimal expansion of the integer that has (1 0101 1111)2 as its binary expansion? Solution: (1 0101 1111)2 = 1∙ 28 + 0∙ 27 + 1∙ 26 + 0∙ 25 + 1∙ 24 + 1∙ 23 + 1∙ 22 + 1∙ 21 + 1∙ 20 =351. Example: What is the decimal expansion of the integer that has (11011)2 as its binary expansion? Solution: (11011)2 = 1 ∙ 24 + 1∙ 23 + 0∙ 22 + 1∙ 21 + 1∙ 20 =27.

Octal Expansions The octal expansion (base 8) uses the digits {0, 1, 2, 3, 4, 5, 6, 7}. Example: What is the decimal expansion of the number with octal expansion (7016)8 ? Solution: 7∙ 83 + 0∙ 82 + 1∙ 81 + 6∙ 80 =3598 Example: What is the decimal expansion of the number with octal expansion (111)8 ? Solution: 1∙ 82 + 1∙ 81 + 1∙ 80 = 64 + 8 + 1 = 73

Hexadecimal Expansions The hexadecimal expansion needs 16 digits, but our decimal system provides only 10. So letters are used for the additional symbols. The hexadecimal system uses the digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}. The letters A through F represent the decimal numbers 10 through 15. Example: What is the decimal expansion of the number with hexadecimal expansion (2 AE 0 B)16 ? Solution: 2∙ 164 + 10∙ 163 + 14∙ 162 + 0∙ 161 + 11∙ 160 =175627 Example: What is the decimal expansion of the number with hexadecimal expansion (E 5)16 ? Solution: 1∙ 162 + 14∙ 161 + 5∙ 160 = 256 + 224 + 5 = 485

Base Conversion To construct the base b expansion of an integer n: • Divide n by b to obtain a quotient and remainder. n = bq 0 + a 0 0 ≤ a 0 ≤ b • The remainder, a 0 , is the rightmost digit in the base b expansion of n. Next, divide q 0 by b. q 0 = bq 1 + a 1 0 ≤ a 1 ≤ b • The remainder, a 1, is the second digit from the right in the base b expansion of n. • Continue by successively dividing the quotients by b, obtaining the additional base b digits as the remainder. The process terminates when the quotient is 0. continued →

Algorithm: Constructing Base b Expansions procedure base b expansion(n, b: positive integers with b > 1) q : = n k : = 0 while (q ≠ 0) ak : = q mod b q : = q div b k : = k + 1 return(ak-1 , …, a 1, a 0) • q represents the quotient obtained by successive divisions by b, starting with q = n. • The digits in the base b expansion are the remainders of the division given by q mod b. • The algorithm terminates when q = 0 is reached.

Base Conversion Example: Find the octal expansion of (12345)10 Solution: Successively dividing by 8 gives: • 12345 = 8 ∙ 1543 + 1 • 1543 = 8 ∙ 192 + 7 • 192 = 8 ∙ 24 + 0 • 24 = 8 ∙ 3 + 0 • 3 =8∙ 0+3 The remainders are the digits from right to left , yielding (30071)8.

Comparison of Hexadecimal, Octal, and Binary Representations Initial 0 s are not shown Each octal digit corresponds to a block of 3 binary digits. Each hexadecimal digit corresponds to a block of 4 binary digits. So, conversion between binary, octal, and hexadecimal is easy.

Conversion Between Binary, Octal, and Hexadecimal Expansions Example: Find the octal and hexadecimal expansions of (11111010111100)2. Solution: • To convert to octal, we group the digits into blocks of three (011 111 010 111 100)2, adding initial 0 s as needed. The blocks from left to right correspond to the digits 3, 7, 2, 7, and 4. Hence, the solution is (37274)8. • To convert to hexadecimal, we group the digits into blocks of four (0011 1110 1011 1100)2, adding initial 0 s as needed. The blocks from left to right correspond to the digits 3, E, B, and C. Hence, the solution is (3 EBC)16.

Primes and Greatest Common Divisors Section 4. 3

Primes Definition: A positive integer p greater than 1 is called prime if the only positive factors of p are 1 and p. A positive integer that is greater than 1 and is not prime is called composite. Example: The integer 7 is prime because its only positive factors are 1 and 7, but 9 is composite because it is divisible by 3.

The Fundamental Theorem of Arithmetic Theorem: Every positive integer greater than 1 can be written uniquely as a prime or as the product of two or more primes where the prime factors are written in order of nondecreasing size. Examples: • • 100 = 2 ∙ 5 ∙ 5 = 22 ∙ 52 641 = 641 999 = 3 ∙ 3 ∙ 37 = 33 ∙ 37 1024 = 2 ∙ 2 ∙ 2 ∙ 2 = 210

Conjectures about Primes • Even though primes have been studied extensively for centuries, many conjectures about them are unresolved, including: • Goldbach’s Conjecture: Every even integer n, n > 2, is the sum of two primes. It has been verified by computer for all positive even integers up to 1. 6 ∙ 1018. The conjecture is believed to be true by most mathematicians. • The Twin Prime Conjecture: The twin prime conjecture is that there are infinitely many pairs of twin primes. Twin primes are pairs of primes that differ by 2. Examples are 3 and 5, 5 and 7, 11 and 13, etc. The current world’s record for twin primes (as of mid 2011) consists of numbers 65, 516, 468, 355∙ 2333, 333 ± 1, which have 100, 355 decimal digits.

Greatest Common Divisor Definition: Let a and b be integers, not both zero. The largest integer d such that d | a and also d | b is called the greatest common divisor of a and b. The greatest common divisor of a and b is denoted by gcd(a, b). One can find greatest common divisors of small numbers by inspection. Example: What is the greatest common divisor of 24 and 36? Solution: gcd(24, 26) = 12 Example: What is the greatest common divisor of 17 and 22? Solution: gcd(17, 22) = 1

Greatest Common Divisor Definition: The integers a and b are relatively prime if their greatest common divisor is 1. Example: 17 and 22 Definition: The integers a 1, a 2, …, an are pairwise relatively prime if gcd(ai, aj)= 1 whenever 1 ≤ i<j ≤n. Example: Determine whether the integers 10, 17 and 21 are pairwise relatively prime. Solution: Because gcd(10, 17) = 1, gcd(10, 21) = 1, and gcd(17, 21) = 1, 10, 17, and 21 are pairwise relatively prime. Example: Determine whether the integers 10, 19, and 24 are pairwise relatively prime. Solution: Because gcd(10, 24) = 2, 10, 19, and 24 are not pairwise relatively prime.

Finding the Greatest Common Divisor Using Prime Factorizations • Suppose the prime factorizations of a and b are: where each exponent is a nonnegative integer, and where all primes occurring in either prime factorization are included in both (so some exponents may be 0). Then: • This formula is valid since the integer on the right (of the equals sign) divides both a and b. No larger integer can divide both a and b. Example: 120 = 23 ∙ 5 500 = 22 ∙ 53 gcd(120, 500) = 2 min(3, 2) ∙ 3 min(1, 0) ∙ 5 min(1, 3) = 22 ∙ 30 ∙ 51 = 20 • Finding the gcd of two positive integers using their prime factorizations is not efficient because there is no efficient algorithm for finding the prime factorization of a positive integer.

Least Common Multiple Definition: The least common multiple of the positive integers a and b is the smallest positive integer that is divisible by both a and b. It is denoted by lcm(a, b). • The least common multiple can also be computed from the prime factorizations. This number is divided by both a and b and no smaller number is divided by a and b. Example: lcm(233572, 2433) = 2 max(3, 4) 3 max(5, 3) 7 max(2, 0) = 24 35 72 • The greatest common divisor and the least common multiple of two integers are related by: Theorem 5: Let a and b be positive integers. Then ab = gcd(a, b) ∙lcm(a, b) (proof is Exercise 31)

Euclidean Algorithm Euclid (325 B. C. E. – 265 B. C. E. ) • The Euclidian algorithm is an efficient method for computing the greatest common divisor of two integers. It is based on the idea that gcd(a, b) is equal to gcd(a, c) when a > b and c is the remainder when a is divided by b. Example: Find gcd(91, 287): • 287 = 91 ∙ 3 + 14 • 91 = 14 ∙ 6 + 7 • 14 = 7 ∙ 2 + 0 Divide 287 by 91 Divide 91 by 14 Divide 14 by 7 Stopping condition gcd(287, 91) = gcd(91, 14) = gcd(14, 7) = 7 continued →

Euclidean Algorithm • The Euclidean algorithm expressed in pseudocode is: procedure gcd(a, b: positive integers) x : = a y : = b while y ≠ 0 r : = x mod y x : = y y : = r return x {gcd(a, b) is x} • In Section 5. 3, we’ll see that the time complexity of the algorithm is O(log b), where a > b.

Correctness of Euclidean Algorithm Lemma 1: Let a = bq + r, where a, b, q, and r are integers. Then gcd(a, b) = gcd(b, r). Proof: • Suppose that d divides both a and b. Then d also divides a − bq = r (by Theorem 1 of Section 4. 1). Hence, any common divisor of a and b must also be any common divisor of b and r. • Suppose that d divides both b and r. Then d also divides bq + r = a. Hence, any common divisor of b and r must also be a common divisor of a and b. • Therefore, gcd(a, b) = gcd(b, r).

Correctness of Euclidean Algorithm r 0 = r 1 q 1 + r 2 • Suppose that a and b are positive r 1 = r 2 q 2 + r 3 integers with a ≥ b. ∙ Let r 0 = a and r 1 = b. ∙ Successive applications of the division ∙ rn-2 = rn-1 qn-1 + r 2 algorithm yields: rn-1 = rnqn. 0 ≤ r 2 < r 1 , 0 ≤ r 3 < r 2 , 0 ≤ rn < rn-1, • Eventually, a remainder of zero occurs in the sequence of terms: a = r 0 > r 1 > r 2 > ∙ ∙ ∙ ≥ 0. The sequence can’t contain more than a terms. Finiteness • By Lemma 1 gcd(a, b) = gcd(r 0, r 1) = ∙ ∙ ∙ = gcd(rn-1, rn) = gcd(rn , 0) = rn. • Hence the greatest common divisor is the last nonzero remainder in the sequence of divisions. Correctness